The pressure of a gas enclosed in a rigid container quadruples. By what factor does its temperature change?

in a titration 20.0 mL of 0.150 M NaOH(aq) Solution exactly neutralize is 24.0 mL of an HCl solution what is the concentration o

dusya [7]

Answer:

0.125

Explanation:

i took a test and got it right choosing this answer hopes this help

5 0

3 years ago

PLEASE HELP!!!! THIS IS DUE AT 12!!!!!

Nina [5.8K]

Answer:

.

Explanation:

3 0

3 years ago

How much heat do you need to raise the temperature of 100 g of aluminum from 30°C to 150°C?

BaLLatris [955]

Heat energy can be calculated by using the specific heat of a substance multiplying it to the mass of the sample and the change in temperature. It is expressed as:

<span>Energy = mCΔT</span>

<span>Energy = 100 (0.90) (150 - 30)</span>

Energy = 10800 J

3 0

4 years ago

Read 2 more answers

Be sure to answer all parts.

MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol$

Now, molarity of ethanol solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M$

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol$

Now, molarity of sucrose solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M$

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol$

Now, molarity of sodium chloride solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M$

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

4 0

3 years ago

Predict the signs of !iH, !).S, and !).G of the system for the following processes at 1 atm: (a) ammonia melts at - 60°C, (b) am

Elza [17]

Answer:

Explanation:

(a) Given,

Normal melting point of ammonia is -77.7 °C

In this process , ammonia melts at - 60°C

Now , during the process of melting , ammonia in the solid phase changes to liquid phase, and this increases the number of particles , therefore , the sign of entropy would be positive . And since , the process is an endothermic process , hence , the change in enthalphy sign will be positive .

Even , the temperature of the ammonia is more than its normal melting point .

Therefore ,

TΔS > ΔH

Hence ,

ΔG < 0

Therefore ,

ΔH = Positive

ΔS = Positive

ΔG = Negative.

(b)Given,

Normal melting point of ammonia is -77.7 °C

In this process , ammonia melts at -77.7 °C

Now , during the process of melting , ammonia in the solid phase changes to liquid phase, and this increases the number of particles , therefore , the sign of entropy would be positive . And since , the process is neither endothermic process nor exothermic process, hence , the change in enthalphy sign will be neutral / zero .

Hence ,

ΔG < 0

Therefore ,

ΔH = 0

ΔS = Positive ΔG = Negative.

(c) Given,

Normal melting point of ammonia is -77.7 °C

In this process , ammonia melts at - 100°C

Now , during the process of melting , ammonia in the solid phase changes to liquid phase, and this increases the number of particles , therefore , the sign of entropy would be positive . And since , the process is an endothermic process , hence , the change in enthalphy sign will be positive .

Even , the temperature of the ammonia is less than its normal melting point .

Hence ,

ΔG > 0

Therefore ,

ΔH = Positive

ΔS = Positive

ΔG = Negative.

7 0

4 years ago